$A$ wire of length $L$ carries a current $I$. If the wire is turned into a square coil of single turn,the maximum magnitude of the torque in a given magnetic field $\overrightarrow{B}$ is

$A$ rectangular coil has $2n$ turns,an area $2a$ and carries a current $2I$. The plane of the coil is at $60^{\circ}$ to a horizontal uniform magnetic field of flux density $B$. The torque on the coil due to magnetic force is

$A$ circular coil of $30$ turns and radius $8 \,cm$ carrying a current of $6 \,A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \,T$. The field lines make an angle of $20^{\circ}$ with the normal of the coil. The magnitude of the counter torque that must be applied to prevent the coil from turning is (in $\,Nm$)

$A$ square loop of side $l$ is kept in a uniform magnetic field $B$ such that its plane makes an angle $\alpha$ with $\vec{B}$. The loop carries a current $i$. The torque experienced by the loop in this position is

$A$ closely wound solenoid of $2000$ turns and area of cross-section $1.5 \times 10^{-4} \, m^2$ carries a current of $2.0 \, A$. It is suspended through its centre and perpendicular to its length,allowing it to turn in a horizontal plane in a uniform magnetic field of $5 \times 10^{-2} \, T$ making an angle of $30^o$ with the axis of the solenoid. The torque on the solenoid will be:

$A$ current-carrying coil experiences a torque due to a magnetic field. The value of the torque is $80 \%$ of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is